# Thread: Probablility help with combination/permutations

1. ## Probablility help with combination/permutations

Solve for n if nP5 = 120 x nC3

2. Originally Posted by mathamatics112
Solve for n if nP5 = 120 x nC3
$
^nP_r = \frac{n!}{(n-r)!}$

$^nC_r = \frac{n!}{(n-r)!r!}$

$^nP_5 = \frac{n!}{(n-5)!} = n(n-1)(n-2)(n-3)(n-4)$ (eq1)

$120\cdot ^nC_3 = 120\cdot\frac{n!}{(n-3)!3!} = 120\cdot \frac{n(n-1)(n-2)}{6}$ (eq2)

eq1 = eq2

$n(n-1)(n-2)(n-3)(n-4) = 20(n)(n-1)(n-2)$

n(n-1)(n-2) will cancel:

$n^2-7n-8 = 0$

solve to get n = 8 or n = -1.

However since n must be greater than 0 only n = 8 is a solution